Describe the ring $\mathbb{Z}[\sqrt{-6}]/\langle2+\sqrt{-6}\rangle$ as a quotient of $\mathbb{Z}$. I would like to know if I am doing this properly. I am trying to self-learn field theory and Galois theory before next semester starts to be better prepared, and this question is from an old exam. 
First, all of the elements of $\mathbb{Z[\sqrt{-6}]}$ are the form $a+b\sqrt{-6}$ with $a, b \in \mathbb{Z}$. Then 
$$a+b\sqrt{-6} + \langle2 + \sqrt{-6}\rangle = a+b\sqrt{-6} - (2b+b\sqrt{-6}) + \langle2 + \sqrt{-6}\rangle = a-2b + \langle2 + \sqrt{-6}\rangle.$$
Moreover, since $\langle2 + \sqrt{-6}\rangle$ is an ideal of $\mathbb{Z}[\sqrt{-6}]$, it is closed under multiplication, and so $(2+\sqrt{-6})(2-\sqrt{-6}) = -2 \in \langle2 + \sqrt{-6}\rangle. $
This means that we can reduce $a-2b$ modulo $2$, so this is either $0$ or $1$. 
Hence, $\mathbb{Z}[\sqrt{-6}]/\langle2+\sqrt{-6}\rangle \cong \mathbb{Z}_2$? 
 A: You have the right idea, but there are two errors.  First, your work shows that $\mathbb{Z}[\sqrt{-6}]/\langle2+\sqrt{-6}\rangle$ has at most two elements.  To be sure that it is really $\mathbb{Z}_2$ rather than the zero ring $\{0\}$ (or $\mathbb{Z}_1$, if you like), you would need to show that $0+\langle2+\sqrt{-6}\rangle$ and $1+\langle2+\sqrt{-6}\rangle$ are not equal.
Second, $(2+\sqrt{-6})(2-\sqrt{-6})$ is $10$, not $-2$.  So actually, your conclusion should be that $\mathbb{Z}[\sqrt{-6}]/\langle2+\sqrt{-6}\rangle$ has at most $10$ elements (or more precisely, there is a surjective homomorphism $\mathbb{Z}_{10}\to \mathbb{Z}[\sqrt{-6}]/\langle2+\sqrt{-6}\rangle$).
Here's one way to show that it really is $\mathbb{Z}_{10}$, rather than $\mathbb{Z}_5$, $\mathbb{Z}_2$, or $\mathbb{Z}_1$.  It suffices to show that $2\not\in\langle2+\sqrt{-6}\rangle$ and $5\not\in \langle2+\sqrt{-6}\rangle$.  If $2\in\langle2+\sqrt{-6}\rangle$, that means there are $a,b\in\mathbb{Z}$ such that $$2=(2+\sqrt{-6})(a+b\sqrt{-6})=(2a-6b)+(a+2b)\sqrt{-6}$$ so $a=-2b$ and $2=2a-6b=-4b-6b=-10b$ which is impossible.  Similarly, if $5\in\langle2+\sqrt{-6}\rangle$ we would have $5=-10b$ which is again impossible.  (Indeed, this computation shows that the integers in $\langle2+\sqrt{-6}\rangle$ are exactly the multiples of $10$, so two integers give the same coset in $\mathbb{Z}[\sqrt{-6}]/\langle2+\sqrt{-6}\rangle$ iff they differ by a multiple of $10$.  Since every coset can be represented by an integer, this proves more directly that $\mathbb{Z}[\sqrt{-6}]/\langle2+\sqrt{-6}\rangle\cong\mathbb{Z}_{10}$.)
